$\begingroup$@DavidEppstein the base field is $\mathbb{F}_2$.$\endgroup$
– NadoJan 28 '16 at 15:29

$\begingroup$@Kaveh A random $n\!\times\!n$ $M$ matrix of elements in $\mathbb{F}_2$ is non-singular with high probability, so I guess multiplying $M$ by $\begin{pmatrix} I_k & 0\\0&0\end{pmatrix}$$\endgroup$
– NadoJan 28 '16 at 15:32